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 March 23rd, 2011, 06:45 PM #1 Newbie   Joined: Mar 2011 Posts: 1 Thanks: 0 Related Rates Water is draining from a cone-shaped filter at the constant rate of 3 cm^3/second. The cone is 60 cm high and 20 cm in diameter. When the cone is half full, how fast is the depth of the water changing?
 March 23rd, 2011, 06:49 PM #2 Senior Member   Joined: Nov 2010 Posts: 502 Thanks: 0 Re: Related Rates Related rates are classical calculus questions, almost always relying on the chain rule to complete. What work do you have so far? The general start is through a formula (here, the volume would be a good one to work with).
 March 23rd, 2011, 07:19 PM #3 Global Moderator     Joined: Nov 2009 From: Northwest Arkansas Posts: 2,766 Thanks: 4 Re: Related Rates Look at the formula for the volume of a cone (as suggested). Try to express this in terms of one variable alone. I'd choose "height", since the question asks about that dimension. Now differentiate. Don't forget the chain rule. There should be some dh/dt in there... At this point, the calculus is over, and it's simply solving for an unknown. Plug in what you know, and use + - x ÷ ? to get the remaining value!
 March 24th, 2011, 03:00 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 462 Math Focus: Calculus/ODEs Re: Related Rates I wanted to give you some time to work it from the suggestions given before posting a solution. At time t, let V be the volume, r be the radius of the volume of water at the surface, and h be the depth of the water. We will need the formula for the volume of a cone: $V=\frac{1}{3}\pi r^2h$ Also, we should observe that for any height h, $r=\frac{1}{6}h$, thus volume as a function of h alone is: $V=\frac{1}{3}\pi $$\frac{1}{6}h$$^2h=\frac{\pi}{108}h^3$ thus we have: $h=$$\frac{108}{\pi}V$$^{\frac{1}{3}}=3$$\frac{4}{\ pi}V$$^{\frac{1}{3}}$ $\frac{dh}{dV}=$$\frac{4}{\pi}V$$^{-\frac{2}{3}}$$\frac{4}{\pi}$$=$$\frac{4}{\pi V^2}$$^{\frac{1}{3}}$ The first sentence tells us $\frac{dV}{dt}=-3\text{ \frac{cm^3}{s}}$ Using the chain rule, we may state: $\frac{dh}{dt}=\frac{dh}{dV}\cdot\frac{dV}{dt}$ When the cone is half full, $V=\frac{1}{2}\cdot\frac{1}{3}\pi$$10\text{ cm}$$^2$$60\text{ cm}$$=1000\pi\text{ cm^3}$ So, we have: $\frac{dh}{dt}=$$\(\frac{4}{\pi \(1000\pi\text{ cm^3}$$^2}\)^{\frac{1}{3}}\)$$-3\text{ \frac{cm^3}{s}}$$=-\frac{3}{50\sqrt[3]{2}\pi}\:\text{\frac{cm}{s}}\approx-0.0151585634454\text{ \frac{cm}{s}}$ Here's another way to approach it: $V=\frac{1}{3}\pi r^2h=2\pi r^3$ $\frac{dV}{dt}=6\pi r^2\frac{dr}{dt}=\pi r^2\frac{dh}{dt}$ $\frac{dh}{dt}=\frac{1}{\pi r^2}\frac{dV}{dt}$ When the cone is half-full, we have: $1000\pi=2\pi r^3$ $r^3=500$ $r^2=50\sqrt[3]{2}$ thus $\frac{dh}{dt}=\frac{1}{50\sqrt[3]{2}\pi}\frac{dV}{dt}=-\frac{3}{50\sqrt[3]{2}\pi}\:\text{\frac{cm}{s}}$

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